Python 如何将0和1分开？

你好，我有这样的号码：

a = "00001"
b = "00011" 
c = "00111"  
d = "01001"
e = "10001"

a = "00001" => "0000", "1"
b = "00011" => "000", "11"
c = "00111" => "00", "111"
d = "01001" => "0", "1", "00", "1"
e = "10001" => "1", "000", "1"

我想要这样的东西：

a = "00001"
b = "00011" 
c = "00111"  
d = "01001"
e = "10001"

a = "00001" => "0000", "1"
b = "00011" => "000", "11"
c = "00111" => "00", "111"
d = "01001" => "0", "1", "00", "1"
e = "10001" => "1", "000", "1"

我怎样才能拆分这个号码

---

多谢各位

将字符串划分为所需的子列表

def encode(s):
  result, prev = [], ''

  for v in s:
    if not prev or prev[-1] == v:
      # continue current pattern
      prev += v
    else:
      # pattern change
      result.append(prev)
      prev = v

  if prev:
    result.append(prev)
  return result

a = "00001"
b = "00011" 
c = "00111"  
d = "01001"
e = "10001"

for s in [a, b, c, d, e]:
  print(s, encode(s))

输出

00001 ['0000', '1']
00011 ['000', '11']
00111 ['00', '111']
01001 ['0', '1', '00', '1']
10001 ['1', '000', '1']

---

Python的一个优点是有很多方法可以完成相同的任务。下面是另一个使用regex的方法

我扩展了您的示例，以显示它仅适用于0或1

import re

a = "00001"
b = "00011" 
c = "00111"  
d = "01001"
e = "10001"
f = "52255"
g = "90099"
h = "81811"

def parse(s):
    """Parse binary string into like groups."""
    f = re.findall('([0]+)|([1]+)', s)
    return [z if z else o for z, o in f]

for i in [a, b, c, d, e, f, g, h]:
    print(i, parse(i))

输出：

---

groupby

就是为此而设计的

导入itertools
def groupthem（二进制）：
“”“将二进制1和0分别分组。
>>>groupthem（“00001”）
["0000", "1"]
>>>分组他们（“00011”）
["000", "11"]
>>>分组他们（“01001”）
["0", "1", "00", "1"]
"""
groups=itertools.groupby（二进制）
return[“”.加入（组）for u，分组]

检查答案可能重复，它可以根据您的问题进行调整